INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A22

SOME RESULTS FOR SUMS OF THE INVERSES OF BINOMIAL COEFFICIENTS

Feng-Zhen Zhao

Department of Applied Mathematics, Dalian University of Technology, Dalian, 116024, China

Tianming Wang

Department of Applied Mathematics, Dalian University of Technology, Dalian, 116024, China

Received: 7/15/04, Revised: 8/11/05, Accepted: 9/26/05, Published: 10/11/05

Abstract

In this paper, the authors establish some identities involving inverses of binomial coefficients and generalize an identity.

1. Introduction

For convenience, we first give some notation. The binomial coefficients are defined by

\binom{n}{m} = \{ \begin{matrix}\frac{n!}{m!(n − m)!}, n ≥ m, \\ 0, n < m, \end{matrix}

where n and m are nonnegative integers.

Binomial coefficients are classical combinatorial numbers, which play an important role in many subjects such as probability, statistics, and number theory. There are many identities related to binomial coefficients. However, computations involving the inverses of binomial coefficients are often difficult. For previous literature dealing with identities related to the inverses of binomial coefficients, see [1–7]. In this paper, we offer some new identities involving inverses of binomial coefficients. To do so, we shall make use of the following formula (see [2])

\binom{n}{k}^{−1} = (n + 1) ∫_{0}^{1} t^{k}(1 − t)^{n−k}dt.  (1)

In particular, we generalize the well-known identity (due to Euler)

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} = \frac{π^{2}}{18}.  (2)

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2. Main Results In this section, we give the main results of this paper. Theorem Let m be a positive integer. Then

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 − t^{m}(1 − t)^{m}]dt}{t(1 − t)},  (3)
∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 + t^{m}(1 − t)^{m}]dt}{t(1 − t)},  (4)
∑_{n=1}^{∞} \frac{1}{n^{2}(n + 1)\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 − t^{m}(1 − t)^{m}]dt}{t(1 − t)}

+\frac{m}{2} ∫_{0}^{1} \frac{t^{m}(1 − t)^{m} + ln[1 − t^{m}(1 − t)^{m}]}{t^{m+1}(1 − t)^{m+1}}dt, (5)

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}(n + 1)\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 + t^{m}(1 − t)^{m}]dt}{t(1 − t)}

+\frac{m}{2} ∫_{0}^{1} \frac{t^{m}(1 − t)^{m} − ln[1 + t^{m}(1 − t)^{m}]}{t^{m+1}(1 − t)^{m+1}}dt. (6)

Proof. We first prove (3). By definition, we have

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2mn}{mn}} = \frac{m}{2} ∑_{n=1}^{∞} \frac{1}{n(2mn − 1)\binom{2mn−2}{mn−1}}

= \frac{m}{2} ∑_{n=0}^{∞} \frac{1}{(n + 1)(2mn + 2m − 1)\binom{2mn+2m−2}{mn+m−1}}.

It follows from (1) that

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2mn}{mn}} = \frac{m}{2} ∑_{n=0}^{∞} \frac{1}{n + 1} ∫_{0}^{1} t^{mn+m−1}(1 − t)^{mn+m−1}dt.

Noticing that

∑_{n=0}^{∞} \frac{t^{mn+m}(1 − t)^{mn+m}}{n + 1} = − ln[1 − t^{m}(1 − t)^{m}]  (7)

converges uniformly for t ∈ [0, 1], we have (3).

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Now we show that (5) holds.

∑_{n=1}^{∞} \frac{1}{n^{2}(n + 1)\binom{2mn}{mn}} = \frac{m}{2} ∑_{n=1}^{∞} \frac{1}{n(n + 1)(2mn − 1)\binom{2mn−2}{mn−1}}

= \frac{m}{2} ∑_{n=1}^{∞} \frac{1}{n(n + 1)} ∫_{0}^{1} t^{mn−1}(1 − t)^{mn−1}dt

= \frac{m}{2}( ∑_{n=1}^{∞} \frac{∫_{0}^{1} t^{mn−1}(1 − t)^{mn−1}dt}{n} − ∑_{n=1}^{∞} \frac{∫_{0}^{1} t^{mn−1}(1 − t)^{mn−1}dt}{n + 1}).

By using (7), we have

∑_{n=1}^{∞} \frac{1}{n^{2}(n + 1)\binom{2mn}{mn}} = −\frac{m}{2} ∫_{0}^{1} \frac{ln[1 − t^{m}(1 − t)^{m}]dt}{t(1 − t)}

+\frac{m}{2} ∫_{0}^{1} \frac{t^{m}(1 − t)^{m} + ln[1 − t^{m}(1 − t)^{m}]}{t^{m+1}(1 − t)^{m+1}}dt.

The proofs of equalities (4) and (6) follow the same pattern and are omitted. □ To see that (3) is a generalization of (2), let I(a) = −\frac{1}{2} ∫_{0}^{1} \frac{ln[1 − at(1 − t)]dt}{t(1 − t)} (0 ≤ a ≤

1). When a > 0,

I′(a) = \frac{1}{2} ∫_{0}^{1} \frac{dt}{at^{2} − at + 1} = \frac{2}{a}\sqrt{\frac{a}{4 − a}} arctan \sqrt{\frac{a}{4 − a}}.

Then I(a) = 2 (arctan \sqrt{\frac{a}{4−a}} )^{2} + c, where c is a constant. Since lim_{a→0} I(a) = 0, we get I(a) = 2 (arctan \sqrt{\frac{a}{4 − a}} )^{2} . Hence,

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} = I(1) = \frac{π^{2}}{18}.

By computing the integrals in (4)-(6), we can obtain other identities involving inverses of binomial coefficients. For example, if m = 1 in (4), we have

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}\binom{2n}{n}} = −\frac{1}{2} ∫_{0}^{1} \frac{ln[1 + t(1 − t)]dt}{t(1 − t)}.

Put J(a) = −\frac{1}{2} ∫_{0}^{1} \frac{ln[1 + at(1 − t)]dt}{t(1 − t)} (0 ≤ a ≤ 1). When a > 0,

J′(a) = \frac{1}{2} ∫_{0}^{1} \frac{dt}{at^{2} − at − 1} = \frac{1}{\sqrt{a^{2} + 4a}} ln |\frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}}|.

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Using some calculus, we find that

J(a) = −\frac{1}{2}[ ln \frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}}]^{2} + c_{1}.

Since lim_{a→0} J(a) = 0, we have that c_{1} = 0. Thus

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}\binom{2n}{n}} = J(1) = −2( ln \frac{\sqrt{5} − 1}{2})^{2}.  (8)

Let

K(a) = \frac{1}{2} ∫_{0}^{1} \frac{at(1 − t) − ln[1 + at(1 − t)]}{t^{2}(1 − t)^{2}}dt (0 ≤ a ≤ 1).

Then, when 0 < a ≤ 1,

K′(a) = −\frac{a}{2} ∫_{0}^{1} \frac{dt}{at^{2} − at − 1}.

By calculus, we obtain

K′(a) = −\begin{matrix}√ & \overline{a} \\ √ & \overline{a + 4} \end{matrix}[ ln (\sqrt{\frac{a + 4}{a}} − 1) − ln (\sqrt{\frac{a + 4}{a}} + 1)],

K(a) = −\{\frac{\sqrt{a + 4} + \sqrt{a}}{\sqrt{a + 4} − \sqrt{a}} ln (\frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}}) + \frac{\sqrt{a + 4} + \sqrt{a}}{\sqrt{a + 4} − \sqrt{a}} + [ ln (\frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}})]^{2}

−\frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}} ln (\frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}}) + \frac{\sqrt{a + 4} − \sqrt{a}}{\sqrt{a + 4} + \sqrt{a}}\} + c_{2}.

Because lim_{a→0} K(a) = 0, we find that c_{2} = 2. By means of (8), we have

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}(n + 1)\binom{2n}{n}} = −6( ln \frac{\sqrt{5} − 1}{2})^{2} − \sqrt{5} ln (\frac{3 − \sqrt{5}}{2}) − 1.

In the same way, from (5) we obtain

∑_{n=1}^{∞} \frac{1}{n^{2}(n + 1)\binom{2n}{n}} = \frac{\sqrt{3}π}{3} − \frac{π^{2}}{18} − 1.

Acknowledgment The authors wish to thank the anonymous referee for his/her valuable suggestions for this paper.

References

1. Nicolae Pavelescu. “Problem C:1280.” Gaz. Mat. 97.6 (1992): 230.

2. Juan Pla. “The Sum of Inverses of Binomial Coefficients Revisited.” The Fibonacci Quarterly 35.4 (1997): 342–345.

3. Andrew M. Rockett. “Sums of the Inverses of Binomial Coefficients.” The Fibonacci Quarterly 19.5 (1981): 433–437.

4. B. Sury. “Sum of the reciprocals of the Binomial Coefficients.” European J. Combin. 14.4 (1993): 351–353.

5. Tiberiu Trif. “Combinatorial Sums and Series Involving Inverses of Binomial Coefficients.” The Fibonacci Quarterly 38.1 (2000): 79–84.

6. WMC Problems Group. “Problem 10494.” Amer. Math. Monthly 103.1 (1996): 74.

7. B. Sury, Tianming Wang, and Feng-Zhen Zhao. “Identities Involving Reciprocals of Binomial Coefficients.” Journal of Integer Sequences 7(2) (2004): Article 04.2.8.